force

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The 4-force F^\mu = (F^0, \vec{F})\, is defined analogously to that in Newtonian mechanics:

F^\mu = m \frac{d u^\mu}{d\tau} = m \frac{d^2 x^\mu}{d\tau^2}\,.

More generally,

F^\mu = \frac{d p^\mu}{d\tau}\,,

where p^\mu = m\tfrac{d x^\mu}{d\tau}\,.

Relation to 3-force

Noting that p^\mu = (E, \vec{p})\, it is sometimes useful to define the 3-force \vec{f}\, by

\vec{f} = \frac{d \vec p}{dt}\,.

Thus \vec{f} = \tfrac{1}{\gamma} \vec{F}\,. Using the fact that

F^\mu p_\mu = - F^0 p^0 + \vec{F}\cdot \vec{p} = 0\,,

which follows from

\frac{d}{d\tau} (p^2) = -\frac{d}{d\tau}(m^2)=0\,,

(assuming that m\, is constant), we obtain that

F^0 = \tfrac{1}{\gamma m} \vec{F}\cdot\vec{p} = \tfrac{1}{\gamma m} \gamma \vec{f}\cdot\vec{p} = \gamma \vec{f}\cdot\vec{v}\,,

where \vec{v} = \tfrac{d\vec{x}}{dt}\,. Finally

F^\mu = (\gamma \vec{f}\cdot\vec{v}, \gamma \vec{f})\,.

The Newtonian result is that

\frac{d E}{d\tau} = \frac{dt}{d\tau}\frac{dE}{dt} = \gamma \vec{f}\cdot\vec{v}\, implies \frac{dE}{dt} = \vec{f}\cdot\vec{v}\,,

and

\frac{d \vec{p}}{d\tau} = \frac{dt}{d\tau}\frac{d\vec{p}}{dt} = \gamma \vec{f}\, implies \frac{d\vec{p}}{dt} = \vec{f}\,.
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